Gauss's Law and charged paint

In summary, a charged paint of -22.3 µC is spread over the surface of a plastic sphere with a diameter of 10.5 cm. Using the equation E=q/(4πε0r^2), the magnitude of the electric field just inside the paint layer is 0. For (b) and (c), incorrect answers were obtained, possibly due to rounding or incorrect input. The correct answer is likely around -7.27 x 10^7 V/m.
  • #1
yaro99
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0

Homework Statement


A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a charge of -22.3 µC.

(a) Find the magnitude of the electric field just inside the paint layer.
(b) Find the magnitude of the electric field just outside the paint layer.
(c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Homework Equations


$$E=\frac{q}{4\pi\epsilon_0 r^2}$$


The Attempt at a Solution


I got (a) correct (E=0), but I'm getting wrong answers for (b) and (c)

For (b), the radius is 0.0525m, so:
$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.1025)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$

For (c), r=0.0525+0.05=0.1025, so:
$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.1025)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$

EDIT: for some reason my latex code isn't displaying properly, so here are pics of my equations:
EDIT 2: fixed it
 
Last edited:
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  • #2
yaro99 said:

Homework Statement


A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a charge of -22.3 µC.

(a) Find the magnitude of the electric field just inside the paint layer.
(b) Find the magnitude of the electric field just outside the paint layer.
(c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Homework Equations


$$E=\frac{q}{4\pi\epsilon_0 r^2}$$

The Attempt at a Solution


I'm getting wrong answers for (b) and (c)
For (b), the radius is 0.0525m, so:$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.0525)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$
Code:
> (-22.3e-6)/(4*pi*(0.0525^2)*(8.854e-12))
ans =  -7.2717e+07
... hmmm, if this is being computer mediated, then you should check rounding and the exact form of the input, units, stuff like that. Perhaps they'll accept ##\small -7.27\times 10^7##V/m

EDIT: for some reason my latex code isn't displaying properly
That's because you didn't use only latex markup inside the latex tags.
Use the "quote" button (below, right) to see what I did.
 

1. What is Gauss's Law?

Gauss's Law is a fundamental law in physics that relates the electric field to the distribution of electric charges. It states that the electric flux through a closed surface is directly proportional to the net charge enclosed by that surface.

2. How does Gauss's Law apply to charged paint?

Charged paint is a material that contains electrically charged particles. When a charged paint is applied to a surface, the electric field from the charged particles follows Gauss's Law and is proportional to the net charge enclosed by the surface.

3. Can Gauss's Law be used to determine the electric field from charged paint?

Yes, Gauss's Law can be used to determine the electric field from charged paint. By choosing a closed surface around the charged paint and calculating the electric flux through that surface, the electric field can be found.

4. What factors affect the strength of the electric field from charged paint?

The strength of the electric field from charged paint is affected by the amount of charge present in the paint, the distance from the surface of the paint, and the dielectric constant of the medium surrounding the paint.

5. How is Gauss's Law used in practical applications of charged paint?

Gauss's Law is used in practical applications of charged paint to calculate the electric field at different points and determine the direction of the electric field. This information can then be used to control and manipulate the behavior of the charged paint, for example, in electrostatic painting or printing processes.

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